in ∆ABC, A=90°,AB=6cm and AC =2√3cm then find angle B
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In triangle ABC
A = 90°
AB = 6 ; AC=2√3
Tan theta = Perpendicular/Base
Tan B = AC/AB
tan B = 2√3/6
= √3/3
= √3/(√3×√3)
tan B = 1/√3
Angle B = 30°
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A = 90°
AB = 6 ; AC=2√3
Tan theta = Perpendicular/Base
Tan B = AC/AB
tan B = 2√3/6
= √3/3
= √3/(√3×√3)
tan B = 1/√3
Angle B = 30°
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Answered by
2
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